If |x−2|=|x−6| then the value of x is
1
2
3
4
Given |x−2|=|x−6| Squaring both sides
(x−2)2=(x−6)2)
x2−4x+4=x2−12x+36
⇒12x−4x=36−4
8x=32
x=4
If (23)x(32)2x=8116, then x =
The value of λ for which the lines 3x+4y=5, 5x+4y=4 and λx+4y=6 meet at a point is